Home › Community › Questions & Answers › Mental Math Q & A › Two Digit Multiplier in U T method 6671 x 386
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August 28, 2020 at 4:55 am #14388
How to multiply 6671 x 386 using the UT method.
The answer is 2,575,006
i keep getting the answer wrong I’m ok with two digit multiplier using U T method but with three digits I am not doing the formula correctly. It doesn’t explain in the book what the formula is for three digit multiplier also the answer in the book is wrong.
can someone help me please.August 28, 2020 at 6:00 am #14390I would recommend you use the Bridge or the Vitruvian man method for 3 digit multipliers to reduce the cognitive load.
UT Method is recommended for two digit multipliers. However it is still possible to do the UT method with the a 3 digit multiplier.
Here is the illustration of the Method
If you want to multiply 6671 × 386 here are the steps.
Step 1: Add the same number of zeros as the number of digits in the multiplier.
So the multiplier 386 has 3 digits. So we add three zeroes in front of 6671 to get 000671
0 0 0 6 6 7 1 x 386
Step 2: Sum of two pair products is a single digit of the answer.
Use your middle and forefinger to keep track of whether to calculate U or the T. Place each finger in the middle of two pairs.
Step 2.1
Using 6 in 386, the pair product of 0 and 0 is 0 + 0 = 0 (U6T6 = 00 + 00).
Using 8 in 386, the pair product of 0 and 0 is 0 + 0 = 0 (U8T8 = 00 + 00).
Using 3 in 386, the pair product of 0 and 6 is 0 + 1 = 1 (U3T3 = 00 + 18).
Adding the three number together we get 0 + 0 + 1 = 1.U6 T6
= U8 T8
= U3 T3
0 0 0 6 6 7 1 x 386 = 1 _ _ _ _ _ _Step 2.2
Moving to the next set of digits
Using 6 in 386, the pair product of 0 and 0 is 0 + 0 = 0 (U6T6 = 00 + 00).
Using 8 in 386, the pair product of 0 and 6 is 0 + 4 = 4 (U8T8 = 00 + 48).
Using 3 in 386, the pair product of 6 and 6 is 8 + 1 = 9 (U3T3 = 18 + 18).
Adding the three number together we get 0 + 4 + 9 = 13.We attach 3 in 13 as the next digit. But 13 has two digits so we carry over the 1 in 13. So the last digit from the previous step which is 1 becomes 2 (got by carrying over 1+1).
= U6 T6
= U8 T8
= U3 T3
0 0 0 6 6 7 1 x 386 = 2 3 _ _ _ _ _Step 2.3
Moving to the next set of digits
Using 6 in 386, the pair product of 0 and 6 is 0 + 3 = 3 (U6T6 = 00 + 36).
Using 8 in 386, the pair product of 6 and 6 is 8 + 4 = 12 (U8T8 = 48 + 48).
Using 3 in 386, the pair product of 6 and 7 is 8 + 2 = 10 (U3T3 = 18 + 21).
Adding the three number together we get 3 + 12 +10 = 25.We attach 5 in 25 as the next digit. But 25 has two digits, so we carry over the 2 in 25. So the last digit from the previous step which is 3 becomes 5 (got by carrying over 3 + 2).
= U6 T6
= U8 T8
= U3 T3
0 0 0 6 6 7 1 x 386 = 2 5 5 _ _ _ _Step 2.4
Moving to the next set of digits
Using 6 in 386, the pair product of 6 and 6 is 6 + 3 = 9 (U6T6 = 36 + 36).
Using 8 in 386, the pair product of 6 and 7 is 8 + 5 = 13 (U8T8 = 48 + 56).
Using 3 in 386, the pair product of 7 and 1 is 1 + 0 = 1 (U3T3 = 21 + 03).
Adding the three number together we get 9 + 13 +1 = 23.We attach 3 in 23 as the next digit. But 23 has two digits, so we carry over the 2 in 23. So the last digit from the previous step which is 5 becomes 7 (got by carrying over 5 + 2).
= U6 T6
= U8 T8
= U3 T3
0 0 0 6 6 7 1 x 386 = 2 5 7 3 _ _ _Step 2.5
Moving to the next set of digits
Using 6 in 386, the pair product of 6 and 7 is 6 + 4 = 10 (U6T6 = 36 + 42).
Using 8 in 386, the pair product of 7 and 1 is 6 + 0 = 6 (U8T8 = 56 + 08).
Using 3 in 386, the pair product of 1 and _ is 3 + _ = 3 (U3T3 = 03 + –).
Adding the three number together we get 10 + 6 + 3 = 19.We attach 9 in 19 as the next digit. But 19 has two digits, so we carry over the 1 in 19. So the last digit from the previous step which is 3 becomes 4 (got by carrying over 3 + 1).
= U6 T6
= U8 T8
= U3 T3
0 0 0 6 6 7 1 x 386 = 2 5 7 4 9 _ _Step 2.6
Moving to the next set of digits
Using 6 in 386, the pair product of 7 and 1 is 2 + 0 = 2 (U6T6 = 42 + 06).
Using 8 in 386, the pair product of 1 and _ is 8 + _ = 8 (U8T8 = 08 + –).
Using 3 in 386, the pair product of _ and _ is _ + _ = _ (U3T3 = –– + –).
Adding the three number together we get 2 + 8 + _ = 10.We attach 0 in 10 as the next digit. But 10 has two digits, so we carry over the 1 in 10. So the last digit from the previous step which is 49 becomes 50 (got by carrying over 49 + 1).
= U6 T6
= U8 T8
= U3 T3
0 0 0 6 6 7 1 x 386 = 2 5 7 5 0 0 _Step 2.7
Moving to the next set of digits
Using 6 in 386, the pair product of 1 and _ is 6 + _ = 6 (U6T6 = 06 + –).
Using 8 in 386, the pair product of _ and _ is _ + _ = _ (U8T8 = –– + –).
Using 3 in 386, the pair product of _ and _ is _ + _ = _ (U3T3 = –– + –).
Adding the three number together we get 6 + _ + _ = 6.6 becomes the last digit of the answer. .
= U6 T6
= U8 T8
= U3 T3
0 0 0 6 6 7 1 x 386 = 2 5 7 5 0 0 6This is your final answer.
Hope this clarifies.

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